Physics · Oscillations · NEET
No. The formula is T = 2 pi root(L/g). Mass m does not appear anywhere, so a heavy bob and a light bob of the same length swing with exactly the same time period. Reason: a heavier bob feels a bigger restoring force, but it also has more inertia, and the two effects cancel exactly (both depend on m). This is the most common NEET trap.
For small swings (angle below about 10 degrees) the time period does NOT depend on amplitude. This is why pendulum motion is called isochronous. Whether you release it from 2 degrees or 8 degrees, T stays the same. For large angles the small-angle approximation sin(theta) = theta breaks down, and T slowly increases with amplitude, so the motion is no longer perfect SHM.
Because T is proportional to root(L), doubling L multiplies T by root(2), about 1.41 times. To DOUBLE the time period you must make the length 4 times bigger, since T depends on the square root of L. Halving the length divides T by root(2).
T is proportional to 1/root(g), so weaker gravity gives a longer period. On the Moon g is about 1/6 of Earth's, so root(g) is smaller and T becomes about root(6) = 2.45 times larger — the pendulum swings slower. Taking a pendulum to a high mountain or deeper into a mine also changes g slightly and changes T.
Yes, indirectly. Heating the metal rod makes it expand, so length L increases, and since T is proportional to root(L), the time period increases and the clock runs slow. Cooling shrinks L and the clock runs fast. This is why real clocks use temperature-compensated pendulums. In NEET this appears as a length-change problem, not a direct temperature formula.
If the mass of the bob of a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is (x/2) times its original time period. The value of x is:
A pendulum hung from the roof of a tall building moves freely to and fro as a simple harmonic oscillator. The acceleration of the bob is 20 m/s^2 at a distance of 5 m from the mean position. The time period of oscillation is:
Savitha, a Class XI student, performs an experiment to find the effective length L of a simple pendulum. She records the time for 30 oscillations as 60 s. The length she calculates is (take pi^2 = 9.8 and g = 9.8 m/s^2):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Only two: the length L of the pendulum and the acceleration due to gravity g. T = 2 pi root(L/g). Longer length increases T; stronger gravity decreases T.
The mass of the bob, the amplitude (for small angles), and the material of the bob do not affect the time period. These are common decoys in NEET questions.
A larger mass feels a larger restoring force but also has larger inertia. Both effects scale with mass m and cancel exactly, so m drops out of the formula.
Yes. On the Moon g is about 1/6 of Earth's value. Since T is proportional to 1/root(g), the time period becomes about 2.45 times larger, so the pendulum swings slower.
Because T is proportional to root(L), you must make the length 4 times larger to double the time period.
